Hello - I'm interested in converting a 3.3 volt, 3 amp D.C. power
supply to 1.2 volt, 600 mAmp. Is it possible? I looked at a couple
of other posts and used some of the calcs there, but the resistor i
come up with doesn't sound right to me (3.5microOhm, 1260 Watt).
Also, if it is a resistor i'm supposed to use, does it matter which
lead it goes on (+ or -)?
Thanks in advance for the help!
One of the things you _may_ be confused about is that when there are
three variables involved, knowing two of them solves the other one
exactly. Ohm's law is like that, too: E=I*R. Three variables. If
you modify your power supply to put out 1.2 volts, that specifies one
variable, E. Whatever you plan to connect to it, the "load,"
specifies another, R. The third is then a consequence of the other
two and not something you normally design a power supply to limit. In
other words, if you modify the output voltage and hook up your load to
it, the current that flows just happens as a consequence and you don't
need to do much about that except to make sure your power supply is up
to delivering whatever is required.
Now, a lot of "loads" are dynamic, in the sense that they may try to
pull varying amounts of current over time. In these cases, you need
to make sure your modified 1.2 volt power supply is able to adapt and
keep the voltage steady over varying load currents. This is what a
power supply's usual job actually is and why people go to any trouble
at all designing one. So if your load is a varying one, like a radio
may be, then you really do need a "power supply design" of some kind
to hold that 1.2 volts over a varying R.
On the other hand, if you know your load's R very well and it doesn't
change enough to matter, then you could just compute a "dropping
resistor" value and let it go at that. In that case, the result would
look like:
,---------,
| |
(+) | R <-- dropping resistor
,----------, R
| 3.3 volt | |
| power | |
| supply | |
'----------' R <-- 1.2 volt "load"
(-) | R with known "R"
| | or known "I"
'---------'
In this case, you'd compute the dropping resistor value as (3.3 - 1.2)
divided by the load's current or else divided by (1.2 / R), where R is
the load's known resistance. So the computation is either:
(3.3 - 1.2) / I
or,
(3.3 - 1.2) / (1.2 / R)
depending on whether you know the R or the I of the load. Since you
specified 600mA, you might prefer to use the first one I mentioned.
This would work out to:
(3.3V - 1.2V) / 600mA = (3.3V - 1.2V) / 0.6A = 3.5 Ohms
Note that this corresponds vaguely to your own calculation, except
that you wrote "microOhms." That's wrong.
But be wary. It may be that your load specifies a power supply that
is ___capable___ of 600mA. This is quite different that saying that
your load actually ___needs___ 600mA, in practice. It may only
require a typical of 100mA or 200mA and they may have written down
600mA as being "safely sufficient." If that is what is really going
on, computing that single resistor value is probably NOT the way to go
and is certainly going to lead to a wrong calculation. Instead, in
that case, you probably really DO need a proper 1.2 volt power supply
that actually knows how to observe its own output and keep track of
your load's needs as they change from time to time.
In that case, just explain what you are trying to power up as best you
can (as much detail as you can muster) and ask for a specific
solution.
Jon